R中按日期分plot实现Nmin缺失值的linear interpolation插补

完全可以通过现有函数实现需求，不需要手动编写逐点计算的逻辑，核心是使用支持自定义x轴间距的线性插值函数，配合分组操作即可实现按plot独立计算、自动匹配采样日期间隔的插补。

推荐实现方法 #

最简便的方案是用zoo包中的na.approx()函数完成线性插值：该函数支持指定时间变量作为插值的x轴，会自动根据相邻观测点的实际时间差计算线性变化梯度，完全匹配两次采样间隔内Nmin呈线性变化的前提假设，不会默认按等间距采样计算结果。配合dplyr包的分组操作，即可快速实现按plot分组的插补。

实现代码 #

# 加载所需包
library(dplyr)
library(zoo)

# 构造示例数据（修正了原代码中日期转换的格式参数错误，原参数与实际日期格式不匹配会导致日期转换为NA）
df <- data.frame(plot= c(1,2,3,4,5,6,7,8,9,10),
                  date = c("2020-10-01", "2020-10-01","2020-10-01","2020-10-01","2020-10-01","2020-10-01","2020-10-01","2020-10-01","2020-10-01","2020-10-01",
                           "2020-10-08", "2020-10-08","2020-10-08","2020-10-08","2020-10-08","2020-10-08","2020-10-08","2020-10-08","2020-10-08","2020-10-08",
                           "2020-10-29","2020-10-29","2020-10-29","2020-10-29","2020-10-29","2020-10-29","2020-10-29","2020-10-29","2020-10-29","2020-10-29",
                           "2020-11-05","2020-11-05","2020-11-05","2020-11-05","2020-11-05","2020-11-05","2020-11-05","2020-11-05","2020-11-05","2020-11-05"),
                  Nmin = c(100, 120,  50,  60,  70,  80, 100,  70,  30,  50,  90, 130,  60,  60,  60,  90, 105,  60,  25,  40,  NA,  NA,  NA,  NA,  NA,  NA,  
                           NA,  NA,  NA,  NA, 50, 170, 100, 60,  20, 130, 125,  20,   5,   0))
df$date <- as.Date(df$date, format="%Y-%m-%d")
df$Nmin <- as.numeric(df$Nmin)

# 按plot分组插值
df_imputed <- df %>%
  group_by(plot) %>%
  # 每个组内按采样日期升序排列，保证插值顺序正确
  arrange(date, .by_group = TRUE) %>%
  mutate(
    # 指定date为x轴，按实际时间间隔计算线性插补值
    Nmin = na.approx(Nmin, x = date, na.rm = FALSE)
  ) %>%
  ungroup()

注意事项 #

• 插值前必须保证每个plot组内的记录按日期升序排列，否则会出现计算错误
• na.approx()默认对序列首尾位置的缺失值保留NA，如果需要对首尾缺失值做外插，可以调整函数的rule参数
• 如果不想使用dplyr，也可以用基础R的ave()函数配合na.approx()实现分组插值，不需要额外加载数据操作包

以示例数据中plot1的计算结果为例：2020-10-08的Nmin为90，2020-11-05的Nmin为50，两个日期间隔28天，Nmin总变化量为-40，日均变化量约为-1.4286；2020-10-29距离2020-10-08间隔21天，插补值为90 + (-1.4286)*21 = 60，完全符合线性变化的计算逻辑。

内容的提问来源于stack exchange，提问作者Greenfee